in line with Dmitri's answer below: when you say that there is a unique tracial state, I think you mean to say that on a finite factor there is a unique faithful normal tracial state. Otherwise, consider $\ell^\infty$.
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Yemon ChoiNov 24 '10 at 17:52

I think you need to reword your question to be a little more precise. Presumably you are really interested in the case of $\sigma$-finite, properly infinite von Neumann algebras? eom.springer.de/v/v096900.htm
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Yemon ChoiNov 24 '10 at 17:54

On the topic of precision, I assume you want your expectation to be faithful (otherwise any state would do) and normal.
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Martin ArgeramiNov 25 '10 at 13:53

2 Answers
2

Using direct integral decomposition, also known as reduction theory, one can reduce the problem to the case of a factor.
A conditional expectation in this case is a state.
Every factor admits a state, but only σ-finite factors admit faithful states.
Thus if you require the conditional expectation to be faithful, all factors in the direct
integral decomposition must be σ-finite,
otherwise no additional conditions are needed to ensure the existence of a conditional expectation.

The answer is yes, provided that $M$ has a faithful normal semifinite weight (this always exists) that is also semifinite when restricted to the centre (this I'm not so sure how easily can happen).

When $M$ has a faithful normal semifinite weight $\varphi$, with $\varphi|_{Z(M)}$ semifinite, consider the modular group $\sigma_t^\varphi$ associated with $\varphi$. For each $t\in\mathbb{R}$, $\sigma_t^\varphi$ is an automorphism of $M$, and in particular it preserves its centre. This means that
\[
\sigma_t^\varphi(Z(M))=Z(M), \ \ t\in\mathbb{R}
\]
This conditions, by Takesaki's Theorem (IX.4.2 in Takesaki 2, or JFA1972) are equivalent to the existence of a conditional expectation $E:M\to Z(M)$, with $\varphi\circ E=\varphi$. This last condition forces $E$ to be faithful and normal.

You certainly do not need any modular theory to see this.
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Andreas ThomNov 25 '10 at 15:51

1

That wouldn't surprise me, but off the top of my head I wouldn't know how to do it in another way.
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Martin ArgeramiNov 25 '10 at 17:41

1

I think that the main point of Takesaki's theorem is that it characterizes the subalgebras for which a conditional expectation exists. However, if you just want to have a conditional expectation onto the center then why don't you proceed as Dmitri Pavlov in his answer. Of course there are some technicalities hidden in the direct integral decomposition and making measurable choices etc, but that is something you are facing anyway. I think that Takesaki's theorem is difficult (relies on modular theory) in the factor case; but then a direct integral approach has to be followed anyway.
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Andreas ThomNov 25 '10 at 20:44

2

I agree that Takesaki's theorem is difficult, but it is a proven result and it doesn't use direct integrals (I try to avoid them precisely because of the technicalities). In Dmitri's argument, I'm not sure exactly which vN algebras can be decomposed into a direct integral of $\sigma$-finite factors, and I don't immediately see if the expectation obtained is faithful and normal.
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Martin ArgeramiNov 25 '10 at 23:54